Solve the Following Question.(5 Marks)

Question 15 Marks

Solve the following quadratic equations.
$(2+i) x^2-(5-i) x+2(1-i)=0$

Answer

Given equation is
$
(2+i) x^2-(5-i) x+2(1-i)=0
$
Comparing with $a x^2+b x+c=0$, we get
$
a=2+i, b=-(5-i), c=2(1-i)
$
Discriminant $=b^2-4 a c$
$
\begin{aligned}
& =[-(5-i)]^2-4 \times(2+i) \times 2(1-i) \\
& =25-10 i+i^2-8(2+i)(1-i) \\
& =25-10 i+i^2-8\left(2-2 i+i-i^2\right) \\
& =25-10 i-1-8(2-i+1) \ldots . . .\left[i^2=-1\right] \\
& =25-10 i-1-16+8 i-8 \\
& =-2 i
\end{aligned}
$
So, the given equation has complex roots.
These roots are given by
$
\begin{aligned}
x & =\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2-4 \mathrm{ac}}}{2 \mathrm{a} \text { }} \\
& =\frac{-[-(5-\mathrm{i})] \pm \sqrt{-2 \mathrm{i}}}{2(2+\mathrm{i})}=\frac{(5-\mathrm{i}) \pm \sqrt{-2 \mathrm{i}}}{2(2+\mathrm{i})}
\end{aligned}
$
Let $\sqrt{-2 i}=a+b i$, where $a, b \in R$
Squaring on both sides, we get
$
\begin{aligned}
& -2 i=a^2+b^2 i^2+2 a b i \\
& 0-2 i=\left(a^2-b^2\right)+2 a b i \quad \ldots\left[\because i^2=-1\right]
\end{aligned}
$
Equating real and imaginary parts, we get
$
\mathrm{a}^2-\mathrm{b}^2=0 \text { and } 2 \mathrm{ab}=-2
$
$
a^2-b^2=0 \text { and } b=-\frac{1}{a}
$\begin{aligned}
& \therefore \quad x=1-\mathrm{i} \text { or } x=\frac{4}{5}-\frac{2 \mathrm{i}}{5} \\
&
\end{aligned}

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Question 25 Marks

Solve the following quadratic equations.
$x^2-(5-i) x+(18+i)=0$

Answer

Given equation is $x^2-(5-i) x+(18+i)=0$
Comparing with $a x^2+b x+c=0$, we get
$
a=1, b=-(5-i), c=18+i
$
Discriminant $=b^2-4 a c$
$
\begin{aligned}
& =[-(5-i)]^2-4 \times 1 \times(18+i) \\
& =25-10 i+i^2-72-4 i \\
& =25-10 i-1-72-4 i \ldots .\left[\div i^2=-1\right] \\
& =-48-14 i
\end{aligned}
$
So, the given equation has complex roots.
These roots are given by
$
\begin{aligned}
x & =\frac{-\mathrm{h} \pm \sqrt{\mathrm{b}^2-4 a \mathrm{c}}}{2 \mathrm{a}} \text { } \\
& =\frac{-[-(5-\mathrm{i})] \pm \sqrt{-48-14 \mathrm{i}}}{2(1)}=\frac{(5-\mathrm{i}) \pm \sqrt{-48-14 \mathrm{i}}}{2}
\end{aligned}
$
Let $\sqrt{-48-14 i}=a+b i$, where $a, b \in R$
Squaring on both sides, we get
$
\begin{aligned}
& -48-14 \mathrm{i}=\mathrm{a}^2+\mathrm{b}^2 \mathrm{i}^2+2 \mathrm{abi} \\
& -48-14 \mathrm{i}=\left(\mathrm{a}^2-\mathrm{b}^2\right)+2 \mathrm{abi} \quad \ldots\left[\because \mathrm{i}^2--1\right]
\end{aligned}
$
Equating real and imaginary parts, we get $a^2-b^2--48$ and $2 a b=-14$
$
\begin{array}{ll}
\therefore & a^2-b^2=-48 \text { and } b=\frac{-7}{a} \\
\therefore & a^2-\left(\frac{-7}{a}\right)^2=-48 \\
\therefore & a^2-\frac{49}{a^2}=-48 \\
\therefore & a^4-49--48 a^2 \\
\therefore & a^4+48 a^2-49=0 \\
\therefore & \left(a^2+49\right)\left(a^2-1\right)=0 \\
\therefore & a^2=-49 \text { or } a^2=1 \\
& \text { But } a \in \mathbb{R} \\
\therefore & a^2 \neq-49 \\
\therefore & a^2=1 \\
\therefore & a= \pm 1
\end{array}
$
When $a=1, b=\frac{7}{1}=-7$
When $a=-1, b=\frac{-7}{-1}=7$
$
\begin{array}{ll}
\therefore & \sqrt{-48-14 i}- \pm(1-7 i) \\
\therefore & x=\frac{(5-i) \pm(1-7 i)}{2} \\
\therefore & x=\frac{5-i+1-7 i}{2} \text { or } x=\frac{5-i-1+7 i}{2} \\
\therefore & x=3-4 i \text { or } x=2+3 i
\end{array}
$

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Question 35 Marks

Solve the following quadratic equations.
$x^2-(3 \sqrt{ } 2+2 i) x+6 \sqrt{2} i=0$

Answer

Given fecaation is $x^2 \quad \beta \sqrt{2}+20 x+6 \sqrt{2} i=0$ $a-1, b=[3 \sqrt{2}+2 i n c-6 \sqrt{2}$
Discriminant $-b^2$ sac
$-\left[(3 \sqrt{2}+2 m)^2-4 \times 1 \times 6 \sqrt{21}\right.$
- 1B, $12 \sqrt{2} 21+\Delta^2-24 \sqrt{2}$
- $18 \quad 1212 \cdot 4-1 \because r^2-11$
- $14-1212$
5a, the given oquation has compler rook
These roots are given by
Let $\sqrt{4-12 y^2}=-3$ - bi wheie a b e R
Squarire oc bob sidek, ve get
$
\begin{aligned}
& 14-12 \sqrt{2} 1-z^2+1 b^2+2 z^2 \\
& 14-12 \sqrt{2})=\left(a^2-b^2\right)-2 a b i \quad \ldots\left[\left(2 \hat{r}^2=-1\right)\right. \\
&
\end{aligned}
$
Equaire sed ad incgioery parts, wo cet
$
\begin{aligned}
& \text { Whn }=1 \sqrt{2}, b=\frac{-5 \sqrt{2}}{3 \sqrt{2}}=-2 \\
& \text { When a }=-3 \sqrt{2} w=-\frac{-6 \sqrt{2}}{-k \sqrt{2}}-2 \\
& \therefore \quad \sqrt{t+12 \sqrt{2} i}= \pm(3 \sqrt{2}-3) \\
& \therefore \quad x-\frac{0 x y+2 i i-(7 \sqrt{2}-2)}{2} \\
& \therefore \quad x=\frac{0,8+2 i p+12 \sqrt{2}-2)}{2} \\
& \text { ee } x=\frac{(6 \sqrt{4}-2 i j-49 \sqrt{2}-3)}{2} \\
& x-3 \sqrt{2} \text { or } x=2 \\
&
\end{aligned}
$

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Question 45 Marks

Solve the following quadratic equations.
$x^2-(2+i) x-(1-7 i)=0$

Answer

Given equation is $x^2-(2+i) x-(1-7 i)=0$
Comparing with $a x^2+b x+c=0$, we get
$
a=1, b=-(2+i), c=-(1-7 i)
$
Discriminant $=b^2-4 a c$
$
\begin{aligned}
& =[-(2+i)]^2-4 \times 1 \times-(1-7 i) \\
& =4+4 i+i^2+4-28 i \\
& =4+4 i-1+4-28 i \ldots . . .\left[\because i^2=-1\right] \\
& =7-24 i
\end{aligned}
$
So, the given equation has complex roots.
These roots are given by
$
\begin{aligned}
x & =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
& =\frac{-[-(2+i)] \pm \sqrt{7-24 i}}{2(1)} \\
& =\frac{(2+i) \pm \sqrt{7-24 i}}{2}
\end{aligned}
$

Let $\sqrt{7-24 i}=a+b i$, where $a, b \in R$
Squaring on both sides, we get
$
\begin{array}{r}
7-24 i=a^2+i^2 b^2+2 a b i \\
\therefore \quad 7-24 i=\left(a^2-b^2\right)+2 a b i
\end{array}
$
$
\ldots\left[\because i^2=-1\right]
$
Equating real and imaginary parts, we get
$
a^2-b^2-7 \text { and } 2 a b=-24
$
$
\begin{array}{ll}
\therefore & a^2-b^2-7 \text { and } b=\frac{-12}{a} \\
\therefore & a^2-\left(\frac{-12}{a}\right)^4=7 \\
\text { } \\
\therefore & a^2-\frac{144}{a^2}=7 \\
\therefore \quad & a^4-144=7 a^2 \\
\therefore \quad & a^4-7 a^2-144-0 \\
\therefore \quad & \left(a^2-16\right)\left(a^2-9\right)=0 \\
\therefore & a^2=16 \text { or } a^2-9
\end{array}
$
But $a \in R$
$
\begin{array}{ll}
\therefore & a^2 \neq-9 \\
\therefore & a^2=16 \\
\therefore & a= \pm 4
\end{array}
$
When $a-4, b-\frac{-12}{4}=-3$
When $a=-4, b=\frac{-12}{-4}=3$
$
\begin{array}{ll}
\therefore & \sqrt{7-24 i}- \pm(4-3 i) \\
\therefore & x=\frac{(2+i) \pm(4-3)}{2} \text { } \\
\therefore & x-\frac{(2+i)+(4-3)}{2} \text { or } x=\frac{(2+\mathrm{i})-(4-3 \mathrm{i})}{2} \\
\therefore & x=3-\mathrm{i} \text { of } x-1+2 \mathrm{i}
\end{array}
$

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Question 55 Marks

Solve the following quadratic equations.
$x^2+3 i x+10=0$

Answer

Given equation is $x^2+3 i x+10=0$
Comparing with $a \mathrm{x}^2+\mathrm{bx}+\mathrm{c}=0$, we get
$
\begin{aligned}
& a=1, b=3 i, c=10 \\
& \text { Discriminant }=b^2-4 a c \\
& =(3 i)^2-4 \times 1 \times 10 \\
& =9 i^2-40 \\
& =-9-40 \ldots \ldots\left[\because i^2=-1\right] \\
& =-49
\end{aligned}
$
So, the given equation has complex roots.
These roots are given by
$
\begin{aligned}
& x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-3 i \pm \sqrt{-49}}{2(1)} \\
& x=\frac{-3 i \pm 7 i}{2} \text { } \\
& x=\frac{-3 i+7 i}{2} \text { or } x=\frac{-3 i-7 i}{2} \\
& \therefore x=2 i \text { or } x=-5 i
\end{aligned}
$
$\therefore$ the roots of the given equation are $2 \mathrm{i}$ and $-5 \mathrm{i}$.
Check:
$
\begin{aligned}
& \text { If } x=2 i \text { and } x=-5 i \text { satisfy the given equation, then our answer is correct. } \\
& \text { L.H.S. }=x^2+3 i x+10 \\
& =(2 i)^2+3 i(2 i)+10 i \\
& =4 i^2+6 i^2+10 \\
& =10 i^2+10 \\
& =-10+10 \ldots . \cdot\left[i^2=-1\right] \\
& =0 \\
& =\text { R.H.S. } \\
& \text { L.H.S. }=x^2+3 i x+10 \\
& =(-5 i)^2+3 i(-5 i)+10 \\
& =25 i^2-15 i^2+10 \\
& =10 i^2+10 \\
& =-10+10 \ldots\left[-\ldots i^2=-1\right] \\
& =0 \\
& =\text { R.H.S. }
\end{aligned}
$
Thus, our answer is correct.

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Maths (commerce) Solve the Following Question.(5 Marks)

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